Problem: Simplify the following expression and state the condition under which the simplification is valid. $a = \dfrac{k^2 - 64}{k - 8}$
Solution: First factor the polynomial in the numerator. The numerator is in the form ${a^2} - {b^2}$ , which is a difference of two squares so we can factor it as $({a} + {b})({a} - {b})$ $ a = k$ $ b = \sqrt{64} = -8$ So we can rewrite the expression as: $a = \dfrac{({k} {-8})({k} + {8})} {k - 8} $ We can divide the numerator and denominator by $(k - 8)$ on condition that $k \neq 8$ Therefore $a = k + 8; k \neq 8$